Question

In the given figure. AB is the chord of a circle with centre O. AB is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. If ∠ACD = y⁰ and ∠AOD = x⁰, prove that x⁰ = 3y⁰.

Answer 1

Since BC = OB

∠OCB = ∠BOC = y⁰ [∵ Angles opposite to equal sides are equal] 

∠OBA = ∠BOC + ∠OCB = y⁰ + y⁰ = 2y⁰. 

[∵ Exterior angle of a D is equal to the sum of the opposite interior angles] 

Also OA = OB [Radii of the same circle] 

∠OAB = ∠OBA = 2y⁰ [Angles opposite to equal sides of a triangle are equal]

∠AOD = ∠OAC + ∠OCA = 2y⁰ + y⁰ = 3y⁰ 

[∵ Exterior angle of a Δ is equal to the sum of the opposite interior angles]